Find the angle x in the figure below

Question

For reference: In the figure , P and T are points of tangency . Calculate x. (Answer:$90^o$}

Does anyone have any ideas? I couldn't find a way

$PODT$ is cyclic

$\angle PDO \cong \angle DOT\\ \triangle POA :isosceles\\ \triangle OTB: isosceles$

$DO$ is angle bissector $\angle B$

DT = DP

Answer 1

Extend $AP$ and $BT$ to $K$. Then notice that $E$ is an orthocenter of triangle $ABK$ so we only need to prove $D\in KE$.

• Easy angle chase we see that $\angle PDT = 2\angle PKT$ and since $PD = TD$ we see that $D$ is a circum centre for $PETK$ so $D$ halves $KE$ and thus it lies on $KE$.

Answer 2

In this answer I assume that the radius of the circle is 1 and the origin is the centre of the circle.

Drawing tangents from the radiuses at $\theta$ and $\phi$ we get two lines ($DP$ and $DT$ in the diagram) $$ \begin{align} x\cos\theta+y\sin\theta&=1\\ x\cos\phi+y\sin\phi&=1 \end{align} $$ Solving for the $x$ coordinate of the intersection of these two lines, $D$, gives $$ x={\sin\theta-\sin\phi\over \sin(\theta-\phi)}. \tag{1} $$ The line from $(1,0)$ to the radius of $\theta$ ($PB$ in the diagram) has equation $$ (x-1)\cos(90-\theta/2)+y\sin(90-\theta/2)=0 $$ and similarly the line from $(-1,0)$ to the radius of $\phi$ ($AT$ in the diagram) is $$ (x+1)\cos(\phi/2)-y\sin(\phi/2)=0 $$ Solving for the $x$ coordinate of the intersection of these two lines, $E$, gives $$ \begin{align} x&={\cos(\theta/2+\phi/2)\over\cos(\theta/2-\phi/2)}\\ &= {2\sin(\theta/2-\phi/2)\cos(\theta/2+\phi/2) \over 2\sin(\theta/2-\phi/2)\cos(\theta/2-\phi/2)} \\ &= { \sin\theta-\sin\phi \over \sin(\theta-\phi) } \end{align} $$ where I have used the relation $$\sin((a+b)/2)\cos((a-b)/2)=\sin a+\sin b$$ in the numerator.

This is the same $x$ value as in (1), the point of intersection $D$, so the line $ED$ is parallel to the $y$ axis, and thus the angle of the line $ED$ with $AB$, the $x$ axis, is $90^\circ$.

Answer 3

Extend DC and mark D' symmetrical D. M = $ DT \cap AB$. The triangle DMD' is isosceles and as DM = D'M, MC will be the height of the triangle DMD' and so we will have x = 90 degrees